Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int Sy, A, B, T;
int N, Q;
int l[200005], r[200005];
long long sl[200005], sr[200005];
long double work(long double x) {
  int idl = lower_bound(l + 1, l + 1 + N, x) - l;
  int idr = lower_bound(r + 1, r + 1 + N, x) - r;
  if (idl == N + 1 && idr == N + 1)
    return sr[N];
  else if (idl == N + 1)
    return x - l[N] + sr[N - 1];
  else {
    if (l[idl] < r[idr])
      return sr[idr - 1];
    else
      return x - l[idl - 1] + sr[idr - 1];
  }
}
long double solve(long double x, long double y) {
  long double u = (x - A * y / Sy) / (1 - y / Sy);
  long double v = (x - B * y / Sy) / (1 - y / Sy);
  return (work(v) - work(u)) / (v - u) * T;
}
int main() {
  scanf("%d%d%d%d", &Sy, &A, &B, &N);
  T = B - A;
  for (int i = 1; i <= N; i++) {
    scanf("%d%d", &l[i], &r[i]);
    sl[i] = sl[i - 1] + (r[i - 1] - l[i - 1]);
    sr[i] = sr[i - 1] + (r[i] - l[i]);
  }
  int x, y;
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d", &x, &y);
    printf("%.10Lf\n", solve(x, y));
  }
  return 0;
}